For analysis of a calcium sample, it is desirable to obtain a precipitate weighing about 0.50g....
For analysis of a calcium sample, it is desirable to obtain a precipitate weighing about 0.50g. What size sample should be taken from a mixture which is approximately 56% CaO?
Homework Answers
The preipitating reaction is
CaO (s) + CO2 (g) --> CaCO3 (s)
given precipitate mass CaCO3 = 0.5 g
CaCO3 moles = mass / Molar mass of CaCO3
= ( 0.5 g/ 100 g/mol) = 0.005
CaO moles needed = 0.005 ( since 1 CaO gives 1 CaCO3)
CaO mass needed = moles x molar mass of CaO
= 0.005 mol x 56.08 g/mol
= 0.2804 g
given mixture is 56 % which means we have 56 g CaO per 100 g mixture
Hence mass of mixture needed to get 0.2804 g CaO = ( 100 /56) x ( 0.2804 g)
= 0.5 g
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